Constructing Elliptic Curves over Q(T ) with
Moderate Rank
Scott Arms
Department of Mathematics, The Ohio State University, Columbus, OH, 43210
Álvaro Lozano-Robledo
Department of Mathematics, Boston University, Boston, MA 02215
Current address: Department of Mathematics, Cornell University, Ithaca, NY
14853
Steven J. Miller
Department of Mathematics, The Ohio State University, Columbus, Ohio, 43210
Current address: Department of Mathematics, Brown University, Providence, RI
02912
Abstract
We give several new constructions for moderate rank elliptic curves over Q(T ). In
particular we construct infinitely many rational elliptic surfaces (not in Weierstrass
form) of rank 6 over Q using polynomials of degree two in T . While our method generates linearly independent points, we are able to show the rank is exactly 6 without
having to verify the points are independent. The method generalizes; however, the
higher rank surfaces are not rational, and we need to check that the constructed
points are linearly independent.
Key words: elliptic Curves, rational elliptic surfaces, Mordell-Weil rank
1991 MSC: 11G05 (primary), 11G20 (secondary)
Email addresses: arms@math.ohio-state.edu (Scott Arms),
alozano@math.cornell.edu (Álvaro Lozano-Robledo),
sjmiller@math.brown.edu (Steven J. Miller).
Preprint submitted to Elsevier Science
1
Introduction
Consider the elliptic curve E over Q(T ):
y 2 + a1 (T )xy + a3 (T )y = x3 + a2 (T )x2 + a4 (T )x + a6 (T ),
(1.1)
where ai (T ) ∈ Z[T ]. By evaluating these polynomials at integers, we obtain
elliptic curves over Q. By Silverman’s Specialization Theorem, for large t ∈ Z
the Mordell-Weil rank of the fiber Et over Q is at least that of the curve E
over Q(T ).
For comparison purposes, we briefly describe other methods to construct
curves with rank. Mestre [Mes1,Mes2] considers a 6-tuple of integers ai and
Q
defines q(x) = 6i=1 (x − ai ) and p(x, T ) = q(x − T )q(x + T ). There exist polynomials g(x, T ) of degree 6 in x and r(x, T ) of degree at most 5 in
x such that p(x, T ) = g 2 (x, T ) − r(x, T ). Consider the curve y 2 = r(x, T )
over Q(T ). If r(x, T ) is of degree 3 or 4 in x, we obtain an elliptic curve
with points P±i (T ) = (±T + ai , g(±T + ai )). If r(x, T ) has degree 4 we may
need to change variables to make the coefficient of x4 a perfect square (see
[Mor], page 77). Two 6-tuples that work are (−17, −16, 10, 11, 14, 17) and
(399, 380, 352, 47, 4, 0) (see [Na1]). Curves of rank up to 14 over Q(T ) have
been constructed this way, and using these methods Nagao [Na1] has found
an elliptic curve of rank at least 21 and Fermigier [Fe2] one of rank at least 22
over Q. Shioda [Sh2] gives explicit constructions for not only rational elliptic
curves over Q(T ) of rank 2, 4, 6, 7 and 8, but generators of the Mordell-Weil
groups as well, and shows in [Sh1] that 8 is the largest possible rank for a
rational elliptic curve over Q(T ).
We now describe the idea of our method. For E as in (1.1), define
AE (p) =
X
1 p−1
at (p),
p t=0
(1.2)
with at (p) = p + 1 − Nt (p), where Nt (p) is the number of points in Et (Fp ) (we
set at (p) = 0 when p | ∆(t)). Rosen and Silverman [RS] prove a version of
a conjecture of Nagao [Na1] which relates AE (p) to the rank of E over Q(T ).
They show that if E : y 2 = x3 + A(T )x + B(T ), with A(T ), B(T ) ∈ Z[T ], and
Tate’s conjecture (known if E is a rational elliptic surface over Q) holds for E,
then
1 X
−AE (p) log p = rank E(Q(T )).
X→∞ X
p≤X
lim
2
(1.3)
Tate’s Conjecture (for our situation; see [Ta]) states that if L2 (E/Q, s) is
2
the Hasse-Weil L-function of E/Q attached to Hét
(E/Q) and NS(E/Q) is the
Néron-Severi group of E/Q, then L2 (E/Q, s) has a meromorphic continuation
to C and has a pole at s = 2 of order −ords=2 L2 (E/Q, s) = rank NS(E/Q).
An elliptic curve E over Q(T ) is a rational elliptic surface over Q if and only
if one of the following holds:
(1) 0 < max{3deg A(T ), 2deg B(T )} < 12.
(2) 3deg A(T ) = 2deg B(T ) = 12 and ordT =0 T 12 ∆(T −1 ) = 0
(see [Mir,RS]). In this paper we construct special rational elliptic surfaces
where we are able to evaluate AE (p) exactly; see Theorem 1 for a rank 6
example. For these surfaces, we have AE (p) = −r + O( p1 ). By Rosen and
Silverman’s result and the Prime Number Theorem, we can conclude that the
constant r is the rank of E over Q(T ).
The novelty of this approach is that by forcing AE (p) to be essentially constant,
provided E is a rational elliptic surface over Q, we can immediately calculate
the Mordell-Weil rank without having to specialize points and calculate height
matrices. Further, we obtain an exact answer for the rank, and not a lower
bound. Finally, it is often useful to have elliptic curves over Q(T ) with exact
formulas for AE (p); see [Mil2] for applications to lower order density terms
in the Katz-Sarnak Density Conjecture for one-parameter families of elliptic
curves.
If the degrees of the defining polynomials of E are too large, our results are
conditional on Tate’s conjecture if we are able to evaluate AE (p). In many
cases, however, we are unable to evaluate AE (p) to the needed accuracy. Our
method does generate candidate points, which upon specialization yield lower
bounds for the rank. In this manner, curves of rank up to 8 over Q(T ) have
been found.
Modifications of our method may yield curves with higher rank over Q(T ),
though to find such curves requires solving very intractable non-linear Diophantine equations and then specializing the points and calculating the height
matrices to see that they are independent over Q(T ).
For additional constructions, especially for lower rank curves over Q(T ), see
[Fe2]. For a good survey on ranks of elliptic curves, see [RuS]. For applications
of quadratic polynomials to primitive root producing polynomials, see [Moree].
3
2
Constructing Rank 6 Rational Surfaces over Q(T )
2.1 Idea of the Construction
The main idea is as follows: we can explicitly evaluate linear and quadratic
Legendre sums; for cubic and higher sums, we cannot in general explicitly
evaluate the sums. Instead, we have bounds (Hasse, Weil) exhibiting large
cancellation.
The goal is to cook up curves E over Q(T ) where we have linear and quadratic
expressions in T . We can evaluate these expressions exactly by a standard
lemma on quadratic Legendre sums (see Lemma 5 of the appendix for a proof),
which states that if a and b are not both zero mod p and p > 2, then for t ∈ Z
Ã
p−1
X
t=0
at2 + bt + c
p
!

³ ´
(p − 1) a
p
³ ´
=
− a
p
³
0 2
´
if p|(b2 − 4ac)
(2.1)
otherwise.
³ ´
)
= ap , and the t-sum is large.
Thus if p|(b2 − 4ac), the summands are a(t−t
p
Later when we generalize the method we study special curves that are quartic
in T . Let
y 2 = f (x, T )
g(x)
h(x)
DT (x)
=
=
=
=
x3 T 2 + 2g(x)T − h(x)
x3 + ax2 + bx + c, c 6= 0
(A − 1)x3 + Bx2 + Cx + D
g(x)2 + x3 h(x).
(2.2)
Note that DT (x) is one-fourth of the discriminant of the quadratic (in T )
polynomial f (x, T ). When we specialize T to t, we write Dt (x) for one-fourth
of the discriminant of the quadratic (in t) polynomial f (x, t). We will see
that the number of distinct, non-zero roots of the DT (x) control the rank. We
write A − 1 as the leading coefficient of h(x), and not A, to simplify future
computations by making the coefficient of x6 in DT (x) equal A.
Our elliptic curve E is not written in standard form, as the coefficient of x3 is
T 2 −2T +A−1. This is harmless, and later we rewrite the curve in Weierstrass
form. As y 2 = f (x, T ), for the fiber at T = t we have
at (p) = −
X
x(p)
where
³ ´
∗
p
Ã
f (x, t)
p
!
= −
X
x(p)
Ã
is the Legendre symbol. We study −pAE (p) =
4
!
x3 t2 + 2g(x)t − h(x)
,
p
(2.3)
Pp−1 Pp−1 ³f (x,t)´
x=0
t=0
p
.
P
³
´
2ct−D
When x ≡ 0 the t-sum vanishes if c 6≡ 0, as it is just p−1
. Assume
t=0
p
now x 6≡ 0. By the lemma on quadratic Legendre sums (Lemma 5)
Ã
p−1
3 2
X
t=0
!
x t + 2g(x)t − h(x)
p

³ 3´
(p − 1) x
³ ´ p
=
− x3
p
if p|Dt (x)
(2.4)
otherwise.
Our goal is to find integer coefficients a, b, c, A, B, C, D so that DT (x) has six
distinct, non-zero integer roots. We
the roots r1 , . . . , r6³to´be squares in
³ 3want
´
Z, as their contribution is (p − 1) rpi . If ri is not a square, rpi will be 1 for
half the primes and −1 for the other half, yielding no net contribution to the
rank. Thus, for 1 ≤ i ≤ 6, let ri = ρ2i .
Assume we can find such coefficients. Then for large p
−pAE (p) =
=
Ã
p−1
X p−1
X
x=0 t=0
Ã
X p−1
X
x=0 t=0
!
f (x, t)
p
=
Ã
p−1
3 2
X p−1
X
x=0 t=0
!
!
x t + 2g(x)t − h(x)
p
Ã
!
Ã
!
p−1
X f (x, t)
X f (x, t)
X p−1
X
f (x, t)
+
+
p
p
p
x:Dt (x)≡0 t=0
x:xDt (x)6≡0 t=0
X
= 0 + 6(p − 1) −
x:xDt (x)6≡0
Ã
x3
p
!
= 6p.
(2.5)
We must find a, . . . , D such that DT (x) has six distinct, non-zero roots ρ2i :
DT (x) = g(x)2 + x3 h(x)
= Ax6 + (B + 2a)x5 + (C + a2 + 2b)x4 + (D + 2ab + 2c)x3
+ (2ac + b2 )x2 + (2bc)x + c2
= A(x6 + R5 x5 + R4 x4 + R3 x3 + R2 x2 + R1 x + R0 )
= A(x − ρ21 )(x − ρ22 )(x − ρ23 )(x − ρ24 )(x − ρ25 )(x − ρ26 ).
(2.6)
2.2 Determining Admissible Constants a, . . . , D
Because of the freedom to choose B, C, D there is no problem matching coefficients for the x5 , x4 , x3 terms. We must simultaneously solve in integers
2ac + b2 = R2 A
2bc = R1 A
c2 = R0 A.
(2.7)
For simplicity, take A = 64R03 . Then
5
c2 =
64R04
−→
c =
8R02
2bc = 64R03 R1
−→
b =
4R0 R1
2ac + b2 = 64R03 R2
(2.8)
−→ a = 4R0 R2 − R12 .
For an explicit example, take ri = ρ2i = i2 . For these choices of roots,
R0 = 518400, R1 = −773136, R2 = 296296.
(2.9)
Solving for a through D yields
A =
64R03 =
8916100448256000000
c =
8R02 =
2149908480000
b =
4R0 R1 =
−1603174809600
a =
4R0 R2 − R12 =
16660111104
B =
R5 A − 2a =
−811365140824616222208
C =
R4 A − a2 − 2b =
26497490347321493520384
(2.10)
D = R3 A − 2ab − 2c = −343107594345448813363200
We convert y 2 = f (x, T ) to y 2 = F (x, T ), which is in Weierstrass normal form.
We send y → T 2 +2Ty−A+1 , x → T 2 +2Tx−A+1 , and then multiply both sides by
(T 2 + 2T − A + 1)2 . For future reference, we note that
√
√
T 2 + 2T − A + 1 = (T + 1 − A)(T + 1 + A)
= (T − t1 )(T − t2 )
= (T − 2985983999)(T + 2985984001).
(2.11)
We have
f (x, T ) = T 2 x3 + (2x3 + 2ax2 + 2bx + 2c)T − (A − 1)x3 − Bx2 − Cx − D
= (T 2 + 2T − A + 1)x3 + (2aT − B)x2 + (2bT − C)x + (2cT − D)
F (x, T ) = x3 + (2aT − B)x2 + (2bT − C)(T 2 + 2T − A + 1)x
+(2cT − D)(T 2 + 2T − A + 1)2 .
(2.12)
We now study the −pAE (p) arising from y 2 = F (x, T ). It is enough to show
this is 6p+O(1) for all p greater than some p0 . Recall that t1 , t2 are the unique
roots of T 2 + 2T − A + 1 ≡ 0 mod p. We find
6
−pAE (p) =
Ã
p−1
X p−1
X
t=0 x=0
F (x, t)
p
!
Ã
!
X p−1
X F (x, t)
=
t6=t1 ,t2 x=0
p
!
Ã
X p−1
X F (x, t)
+
p
t=t1 ,t2 x=0
.
(2.13)
For
t2 , send x −→ (t2 + 2t − A + 1)x. As (t2 + 2t − A + 1) 6≡ 0,
³ 2 t 6= t1 ,2 ´
(t +2t−A+1)
= 1 and by (2.12) the sum over t 6= t1 , t2 in (2.13) is now of
p
f (x, t) instead of F (x, T ). Simple algebra yields
−pAE (p) =
Ã
!
X p−1
X f (x, t)
p
t6=t1 ,t2 x=0
=
Ã
!
p−1
X p−1
X f (x, t)
t=0 x=0
p
= 6p + O(1) +
Ã
!
X p−1
X x3 + (2at − B)x2 + 0x + 0
+
p
t=t1 ,t2 x=0
+
!
Ã
X p−1
X x + 2at − B
t=t1 ,t2 x=1
p
−
!
Ã
X p−1
X f (x, t)
t=t1 ,t2 x=0
p
Ã
!
X p−1
X (2at − B)x2 + (2bt − C)x + (2ct − D)
t=t1 ,t2 x=0
p
,
(2.14)
where the main term (the 6p) follows from (2.5). By the lemma on quadratic
Legendre sums, the x-sum in (2.14) is negligible (i.e., is O(1)) if
φ(t) = (2bt − C)2 − 4(2at − B)(2ct − D)
(2.15)
is not congruent to zero modulo p when t = t1 or t2 . Calculating yields
φ(t1 ) =
=
φ(t2 ) =
=
4291243480243836561123092143580209905401856
232 · 325 · 75 · 112 · 13 · 19 · 29 · 31 · 47 · 67 · 83 · 97 · 103
4291243816662452751895093255391719515488256
233 · 312 · 7 · 11 · 13 · 41 · 173 · 17389 · 805873 · 9447850813. (2.16)
Hence, except for finitely many primes (coming from factors of φ(ti ), a, . . . , D,
t1 and t2 ), −pAE (p) = 6p+O(1) as desired. We have shown the following result:
Theorem 1 There exist integers a, b, c, A, B, C, D so that the curve E : y 2 =
x3 T 2 + 2g(x)T − h(x) over Q(T ), with g(x) = x3 + ax2 + bx + c and h(x) =
(A−1)x3 +Bx2 +Cx+D, has rank 6 over Q(T ). In particular, with the choices
of a through D above, E is a rational elliptic surface and has Weierstrass form
y 2 = x3 +(2aT −B)x2 +(2bT −C)(T 2 +2T −A+1)x+(2cT −D)(T 2 +2T −A+1)2 .
Proof: We show E is a rational elliptic surface by translating x 7→ x − (2aT −
B)/3, which yields y 2 = x3 + A(T )x + B(T ) with deg(A) = 3, deg(B) = 5.
Therefore the Rosen-Silverman theorem is applicable, and because we can
7
compute AE (p), we know the rank is exactly 6 (and we never need to calculate
height matrices).
2
Remark 2 We can construct infinitely many E over Q(T ) with rank 6 using
(2.10), as for generic choices of roots ρ21 , . . . , ρ26 , (2.15) holds.
For concreteness, we explicitly list a curve of rank at least 6. Doing a better
job of choosing coefficients a through D (but still being crude) yields
Theorem 3 The elliptic curve y 2 = x3 + Ax + B has rank at least 6 over Q,
where
A =
1123187040185717205972
B = 50786893859117937639786031372848.
Six points on the curve are:
(67585071288, 20866449849961716)
(60673071396, 18500949214922664)
(49153071576, 14991664661755236)
(33025071828, 11131001682078096)
(12289072152,
8151425152633980) (−13054927452,
5822267813027064).
(2.17)
As the determinant of the height matrix is approximately 880, 000, the points
are independent and therefore generate the group. A trivial modification of
this procedure yields rational elliptic surfaces of any rank r ≤ 6. For more
constructions along these lines, see [Mil1].
3
More Attempts for Curves with rank 6, 7 and 8 over Q(T )
3.1 Curves of Rank 6
We sketch another construction for a curve of rank 6 over Q(T ) by modifying
our previous arguments. We define a curve E over Q(T ) by
y 2 = f (x, T )
g(x)
h(x)
DT (x)
=
=
=
=
x4 T 2 + 2g(x)T − h(x)
x4 + ax3 + bx2 + cx + d, d 6= 0
−x4 + Ax3 + Bx2 + Cx + D
g(x)2 + x4 h(x).
8
(3.1)
Q
7
2
We must find choices of the free coefficients such³that D
T (x) =
i=1 (α x−ρi ),
P 2dt−D´
with each root non-zero. For x = 0, we have t
= 0. By Lemma 5, for
p
x a root of DT we have a contribution of (p − 1)
³
x4 α−8
p
´
³ 4´
x
p
³
= (p − 1)
ρ4i α−8
p
´
= p − 1;
for all other x a contribution of −
= −1. Hence summing over x and
P
t yields 7(p − 1) + x6=ρi ,0 −1 = 6p. Similar reasoning as before shows we can
find integer solutions (we included the factor of α2 to facilitate finding such
solutions). We chose the coefficient of the x4 term to be T 2 +2T +1 = (T +1)2 ,
as this implies each curve Et is isomorphic over Q to an elliptic curve Et0 (see
Appendix B). As E is almost certainly not rational, the rank is exactly 6 if
Tate’s conjecture is true for the surface. If we only desire a lower bound for
the rank, we can list the 6 points and calculate the determinant of the height
matrix and see if they are independent.
3.2 Probable Rank 7, 8 Curves
We modify the previous construction to
y 2 = x3 T 2 + 2g(x)T − h(x)
g(x) = x4 + ax3 + bx2 + cx + d, d 6= 0
h(x) = Ax4 + Bx3 + Cx2 + Dx + E
(3.2)
to obtain what should be higher rank curves over Q(T ). Choosing appropriate
quartics for g(x), h(x) such that DT (x) = g 2 (x)+x3 h(x) has eight distinct nonzero perfect square roots should yield a contribution of 8p. As the coefficient
of T 2 is x3 , we do not lose p from summing over non-roots of DT (x). By
specializing to T = a2 S 2 +a1 S+a0 for some constants, we can arrange it so y 2 =
k 2 (S)x4 + · · · , and by the previous arguments obtain a cubic. Unfortunately,
we can no longer explicitly evaluate pAE (p) (because of the replacement T →
a2 S 2 + a1 S + a0 ). As the method yields eight points for all s, we need only
specialize and compute the height matrix. As we construct a rank 8 curve
over Q(T ) in §4 (when we generalize our construction), we do not provide the
details here. Note, however, that sometimes there are obstructions and the
rank is lower than one would expect (see §5).
4
Using Cubics and Quartics in T
Previously we used y 2 = f (x, T ), with f quadratic in T . The reason is that,
for special x, we obtain yi2 = si (xi )2 (T − ti )2 . For such x, the t-sum is large
(of size p); we then show for other x that the t-sum is small.
9
4.1 Idea of Construction
The natural generalization of our Discriminant Method is to consider y 2 =
f (x, T ), with f of higher order in T . We first
polynomials cubic
´
³ consider
P
f (xi ,t)
in T . For a fixed xi , we have the t-sum t(p)
, and there are several
p
possibilities:
(1) f (xi , T ) = a(T − t1 )3 . In this case, the t-sum will vanish, as
³
t−t1
p
(2)
(3)
(4)
(5)
´
³
(t−t1 )3
p
´
=
.
2
f (xi , T
³ ) = 2a(T −´ t1 ) ³(T −
´ t2 ). The t-sum will be O(1), as for t 6= t1 we
(t−t1 ) (t−t2 )
t−t2
have
= p .
p
√
f (xi , T ) = a(T − t1 )(T − t2 )(T − t3 ). This will in general be of size p.
f (xi , T ) = a(T − t1 )(T 2 + bT + c), with the quadratic irreducible over
Z/pZ. This happens when b2 − 4c is not a square mod p. This will in
√
general be of size p.
f (xi , T ) = aT 3 + bT 2 + cT + d, with the cubic irreducible over Z/pZ.
√
Again, this will in general be of size p.
Thus, our method ³does not generalize
´ to f (x, T ) cubic in T . The problem is we
(t−t1 )2n1 ···(t−ti )2ni
cannot reduce to
. We therefore investigate f (x, T ) quartic
p
in T . Consider, for simplicity, a curve E over Q(T ) of the form:
y 2 = f (x, T ) = A(x)T 4 + B(x)T 2 + C(x),
(4.3)
A(x), B(x), C(x) ∈ Z[x] of degree at most 4. The polynomial AT 4 + BT 2 + C
has discriminant 16AC(4AC − B 2 )2 . There are several possibilities for special
choices of x giving rise to large t-sums (sums of size p):
(1) A(xi ), B(xi ) ≡ 0 mod p, C(xi ) a non-zero square mod p. Then the tsummand is of the form c2 , contributing p.
(2) A(xi ), C(xi ) ≡ 0 mod p, B(xi ) a non-zero square mod p. Then the tsummand is of the form (bt)2 , contributing p − 1.
(3) B(xi ), C(xi ) ≡ 0 mod p, A(xi ) a non-zero square mod p. Then the tsummand is of the form (at2 )2 , contributing p − 1.
(4) A(xi ) is a non-zero square mod p and B(xi )2 − 4A(xi )C(xi ) ≡ 0 mod p.
Then the t-summand is of the form a2 (t2 − t1 )2 , contributing p − 1.
In the above construction, we are no longer able to calculate AE (p) exactly.
Instead, we construct curves where we believe AE (p) is large. This is accomplished by forcing points to be on E which satisfy any of (1) through (4)
above. As we are unable to evaluate the AE (p) sums, we specialize and calculate height matrices to show the points are independent. Unfortunately, some
of our constructions yielded 9 and 10 points on E, but some of these points
were linearly dependent on the others, or torsion points (see §5).
10
This method, with a quartic in T , can force a maximum number of 12 points
on E. It is possible to have 8 points from the vanishing of the discriminant
(in t), and an additional 6 points from the simultaneous vanishing of pairs
of A(x), B(x), C(x); however, any common root of A or C with B is also a
root of B 2 − 4AC, so there are at most 4 new roots arising from simultaneous
vanishing, for a total of 12 possible points.
4.2 Rank (at least) 7 Curve
For appropriate choices of the parameters, the curve E : y 2 = A(x)T 4 +
4B(x)T 2 + 4C(x) over Q(T ) with
A(x) = a1 a2 a3 a4 (x − a1 )(x − a2 )(x − a3 )(x − a4 )
C(x) = a1 a2 c1 c2 (x − a1 )(x − a2 )(x − c1 )(x − c2 )
B(x) = a21 a22 (x − c1 )(x − c2 )(x − a3 )(x − a4 )
(4.4)
has rank at least 7. We get 6 points from the common vanishing of A, B, C
in pairs and an additional point from a factor of B 2 − AC. Choosing a1 =
−25, a2 = −5, a3 = −10, a4 = −1, c1 = −9, c2 = 15 we find that the points
(−25, 120000T ), (−5, 10000T ), (−10, 11250), (−1, 28800),
(−9, 800T 2 ), (15, 20000T 2 ), (65/7, (540000T 2 − 2880000)/49)
(4.5)
all lie on E. Upon transforming to a cubic (see Appendix B), specializing to
T = 20, and considering the minimal model, we found that these points are
linearly independent (PARI calculates the determinant of the height matrix
is approximately 37472). Note this is not a rational surface, as the coefficient
of x in Weierstrass form is of degree 8.
4.3 Rank (at least) 8 Curve
For appropriate choices of the parameters, the curve E : y 2 = A(x)T 4 +
B(x)T 2 + C(x) over Q(T ) with
A(x) = x4 , B(x) = 2x(b3 x3 +b2 x2 +b1 x+b0 )+b2 , C(x) = x(b23 x3 +c2 x2 +c1 x+c0 )
has rank at least 8. As the coefficient of x4 is T 4 + 2b3 T 2 + b23 , a perfect square,
E can easily be transformed into Weierstrass form (see Appendix B). The
common vanishing of A and C at x = 0 produces a point S0 = (0, bT ) on
E/Q(T ). Also notice that as before, if B 2 − 4AC vanishes at x = xi then we
11
can rewrite:
Ã
B(xi )
A(xi )T + B(xi )T + C(xi ) = A(xi ) T +
2A(xi )
4
2
Thus we obtain a point Pxi
constants bi , b an ci so that
!2
Ã
!2
B(xi )
=
T +
2x4i
(4.6)
= (xi , x2i (T 2 + B(xi )/2x4i )) on E. We chose
2
x4i
2
B 2 − 4AC = (x − 1)(x + 1)(x − 4)(x + 4)(x − 9)(x + 9)(x − 16),
(4.7)
and obtain a curve E over Q(T ) with coefficients:
5852770213 4 89071 3 89233 2 9
x +
x −
x − x + 144,
382205952
36864
1152
2
34254919166180065369 4 528356915749387 3
C(x) =
x −
x
584325558976905216
28179280429056
527067904642903 2 5881576729
+
x −
x.
(4.8)
880602513408
169869312
A = x4 ,
B(x) = −
As discussed above, the curve E given by (4.8) has 8 rational points over Q(T ),
namely S0 and Pxi for xi = ±1, ±4, ±9, 16. As E is not a rational surface,
and as we cannot evaluate AE (p) exactly, we need to make sure the points
are linearly independent. Specializing to T = 1 yields the elliptic curve with
minimal model
E1 : y 2 =
x3 − x2 − αx + β
α =
357917711928106838175050781865
(4.9)
β = 8790806811671574287759992288018136706011725.
The eight points of ET at T = 1 are linearly independent on E1 /Q (PARI
calculates the determinant of the height matrix to be about 124079248627.08),
proving E does have rank at least 8 over Q(T ).
5
Linear Dependencies Among Points
Not all choices of A(x), B(x), C(x) which yield r points on the curve E : y 2 =
A(x)T 4 + 4B(x)T 2 + 4C(x) actually give a curve of rank at least r over Q(T ).
We found many examples giving 9 and 10 points by choosing A(x) = C(x) so
that B 2 − AC factors nicely, and then searching through prospective roots of
this quantity as well as roots of A(x) = C(x). One such curve giving 10 points
arises from
12
A(x) = C(x) = (x − 1)2 (2x − 1)2
B(x) = 12316x4 + 2346x3 − 239x2 − 24x + 1,
(5.10)
and has the following points on it
µ
¶
µ
¶
−1 420 2
−1 15 2
,
(T + 2) ,
, (T + 2) ,
19 361
4 8
µ
¶ µ
¶ µ
¶
1 56 2
−1 72 2
−1 42 2
, (T + 2) ,
, (T − 2) ,
, (T − 2) ,
9 81
7 49
5 25
µ
¶ µ
¶
µ
¶
1 90 2
1 105 2
1
,
(T − 2) ,
,
(T − 2) , (1, 240T ),
, 63T .
11 121
16 128
2
(0, T 2 + 2),
(5.11)
It can be shown, however, that upon translating to a cubic only the (translated versions of the) second, third, fifth, sixth, and ninth of these points are
independent over Q(T ). While the contribution from these points makes AE (p)
want to be large, this is not reflected by a large rank.
6
Using Higher Degree Polynomials
Let f (x, T ) be a polynomial of degree 3 or 4 in x and arbitrary degree in T
and let E be the elliptic curve over Q(T ) given by y 2 = f (x, T ) (with the
coefficient of x4 a perfect square or zero). The remarks at the beginning of
Section 4 about cubics suggest that we should look for polynomials f (x, T )
with even degree in T , say degT (f ) = 2n.
The nice feature of quadratics and biquadratics that we used in the previous
constructions was the fact that a zero of the discriminant indicates that the
polynomial f (x, T ) factors as a perfect square. However, when f is of arbitrary degree 2n in T this is no longer true: a zero of the discriminant DT (x)
indicates just a multiple root. However, in the most general case, there exist
n quantities Di,T (x) such that their common vanishing at x = x0 implies that
f (x, T ) factors as a perfect square. As an example we look at a quartic of the
form f (x, T ) = A2 T 4 + BT 3 + CT 2 + DT + E 2 , where degx (A, E) ≤ 2 and
degx (B, C, D) ≤ 4. This can be rewritten as:
Ã
2
AT
4
+ 2AT
Ã
+
B
D−
A
2
Ã
Bt
C
B2
+
−
2A 2A 8A3
B2
C
−
2A 8A3
!
!!
T −
13
Ã
+
Ã
C
B2
BT
+
−
2A
2A 8A3
B2
C
−
2A 8A3
!2
!2
+ E 2.
(6.12)
The last two terms are the ones which are keeping the polynomial from being
a perfect square. Thus, if
B
D−
A
Ã
C
B2
−
2A 8A3
!
Ã
= 0,
C
B2
E2 −
−
2A 8A3
!2
= 0
(6.13)
then the polynomial f will be a square. This is equivalent to
D1,T = 8A4 D − 4A2 BC + B 3 = 0
D2,T = 64A6 E 2 − 16A4 C 2 − B 4 + 8A2 CB 2 = 0.
(6.14)
Note that if B=D=0, the conditions that these polynomials impose reduce
to the usual discriminant. Also, degx (D1,T ) ≤ 12, degx (D2,T ) ≤ 16, so we
could get up to 12 points of common vanishing of the Di . The authors have
tried to find suitable constants without success, due to the complexity of the
Diophantine equations.
Acknowledgements
We thank Vitaly Bergelson, Michael Hunt, James Mailhot, David Rohrlich,
Mustafa Sahin, and Warren Sinnott for many enlightening conversations, and
the referee for a very careful reading of the manuscript. The third named
author also wishes to thank Boston University for its hospitality, where much
of the final write-up was prepared.
A
Sums of Legendre Symbols
For completeness, we provide proofs of the quadratic Legendre sums that are
used in our constructions.
A.1 Factorizable Quadratics in Sums of Legendre Symbols
Lemma 4 For p > 2
S(n) =
Ã
p−1
X
x=0
n1 + x
p
!Ã
n2 + x
p
!
=
14

p − 1
−1
if p|(n1 − n2 )
otherwise.
(A.1)
Proof: Translating x by −n2 , we need only prove the lemma when n2 = 0.
Assume (n, p) = 1 as otherwise the result is trivial. For (a, p) = 1 we have:
S(n) =
=
=
Ã
p−1
X
n+x
p
!Ã !
x=0
Ã
p−1
X
x
p
n + a−1 x
p
x=0
Ã
p−1
X
an + x
p
x=0
!Ã
a−1 x
p
!
!Ã !
x
p
= S(an)
(A.2)
Hence
Ã
!Ã !
Ã
!Ã !
X p−1
X an + x
1 p−1
S(n) =
p − 1 a=1 x=0
p
X p−1
X an + x
1 p−1
=
p − 1 a=0 x=0
p
à !
Ã
x
p
à !2
X x
x
1 p−1
−
p
p − 1 x=0 p
!
X x p−1
X an + x
1 p−1
−1
=
p − 1 x=0 p a=0
p
= 0 − 1 = −1.
(A.3)
2
³
P
´
an+x
We need p > 2 as we used p−1
= 0 for (n, p) = 1. This is true for all
a=0
p
p−1
odd primes (as there are 2 quadratic residues, p−1
non-residues, and 0); for
2
p = 2, there is one quadratic residue, no non-residues, and 0.
A.2 General Quadratics in Sums of Legendre Symbols
Lemma 5 (Quadratic Legendre Sums) Assume a and b are not both zero
mod p and p > 2. Then
Ã
p−1
X
t=0
at2 + bt + c
p
!

³ ´
(p − 1) a
p
³ ´
=
− a
p
if p|(b2 − 4ac)
otherwise.
(A.4)
Proof: Assume a 6≡ 0(p) ³as otherwise
the proof is trivial. By translating t, we
´
P
t2 −δ
reduce to the case t(p) p , where δ = b2 − 4ac is the discriminant. If p|δ,
the claim is clear. For p 6 |δ the claim is equivalent to counting the number of
solutions to t2 − δ ≡ y 2 mod p, or (t − y)(t + y) ≡ δ mod p. Letting u = t − y
15
and v = t + y we see there are p − 1 pairs (u, v) with δ ≡ uv mod p (as
δ 6≡ 0). Using that the pairs (u, v) are in bijection with the pairs³(t,´y), the
proof is then easily completed on distinguishing between the case −δ
= −1
p
and
³
−δ
p
´
= 1.
2
Proof: Assume a 6≡ 0(p) as otherwise the proof is trivial. Let δ = 4−1 (b2 −4ac).
Then
Ã
p−1
X
t=0
at2 + bt + c
p
!
=
=
=
Ã
!Ã
p−1
−1
2 2
X
t=0
Ã
p−1
X
t=0
Ã
p−1
X
t=0
=
a
p
!Ã
a t + bat + ac
p
!
!
a
p
t2 + bt + ac
p
a
p
(t + 2−1 b)2 − 4−1 (b2 − 4ac)
p
!Ã
!
à ! p−1 Ã
!
a X t2 − δ
p
(A.5)
p
t=0
If δ ≡ 0(p) we get p − 1. If δ ≡ η 2 , η 6= 0, then by Lemma 4
Ã
p−1
X 2
t −δ
p
t=0
P
³
2
!
=
Ã
p−1
X
t=0
t−η
p
!Ã
t+η
p
!
= −1.
(A.6)
´
t −δ
We note that p−1
is the same for all non-square δ’s (let g be a generator
t=0
p
of the multiplicative group, δ = g 2k+1 , change variables by t → g k t). Denote
this sum by S, the set of ³non-zero
squares mod p by R, and the non-squares
´
P
t2 −δ
mod p by N . Since p−1
=
0
we have
δ=0
p
Ã
p−1
X p−1
X 2
δ=0 t=0
t −δ
p
!
=
à !
p−1
X 2
t=0
t
p
Ã
!
X p−1
X t2 − δ
+
= (p − 1) +
δ∈R t=0
+
δ∈N t=0
p−1
p−1
(−1) +
S = 0
2
2
Hence S = −1, proving the lemma.
B
p
Ã
!
X p−1
X t2 − δ
p
(A.7)
2
Converting from Quartics to Cubics
We record two useful transformations from quartics to cubics. In all theorems
below, all quantities are rational.
16
Theorem 6 If the quartic curve y 2 = x4 − 6cx2 + 4dx + e has a rational point,
then it is equivalent to the cubic curve Y 2 = 4X 3 − g2 X − g3 , where
g2 = e + 3c2 ,
g3 = −ce − d2 + c3 ,
(B.1)
and
2x = (Y − d)/(X − c),
y = −x2 + 2X + c.
(B.2)
See [Mor], page 77. Note that if the leading term of the quartic is a2 x4 , one
can send y → y/a and x → x/a.
Theorem 7 The quartic v 2 = au4 + bu3 + cu2 + du + q 2 is equivalent to the
cubic y 2 + a1 xy + a3 y = x3 + a2 x2 + a4 x + a6 , where
a1 = d/q, a2 = c − (d2 /4q 2 ), a3 = 2qb, a4 = −4q 2 a, a6 = a2 a4
(B.3)
and
x =
2q(v + q) + du
,
u2
y =
4q 2 (v + q) + 2q(du + cu2 ) − (d2 u2 /2q)
. (B.4)
u3
The point (u, v) = (0, q) corresponds to (x, y) = ∞ and (u, v) = (0, −q)
corresponds to (x, y) = (−a2 , a1 a2 − a3 ).
See [Wa], page 37.
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18